The U-shaped glass tube in the figure below contains two liquids in mechanical equilibrium: sea water of density 1.025 kg/L, and an unknown liquid of density ρ. The unknown liquid is in the left tube, floating on top of the water with a clearly visible interface. Use h1 = 14.6 cm and h2 = 1.0 cm with the heights as labeled in the figure below. What is the density of the unknown liquid?
What question?
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To determine the density (ρ) of the unknown liquid, we can use the principle of mechanical equilibrium.
In mechanical equilibrium, the pressure at any given depth is the same for both liquids in the U-shaped tube.
Let's denote the densities of sea water and the unknown liquid as ρ1 and ρ2 respectively.
The pressure at any given depth is given by the equation:
P = ρ * g * h
where P is the pressure, ρ is the density, g is the acceleration due to gravity, and h is the height (depth).
At the interface between the two liquids, the pressure is the same. Therefore, we can equate the pressures on both sides to find the density of the unknown liquid:
ρ1 * g * h1 = ρ2 * g * h2
Substituting the given values:
1.025 kg/L * 9.8 m/s^2 * 0.146 m = ρ2 * 9.8 m/s^2 * 0.01 m
Simplifying:
0.151 kg/m^3 = ρ2
Therefore, the density of the unknown liquid is approximately 0.151 kg/m^3.
To find the density of the unknown liquid, we can use the concept of pressure equilibrium.
First, let's consider the pressure at the interface between the two liquids. The pressure at that point is the same, since the system is in mechanical equilibrium. The pressure at a certain depth in a liquid can be calculated using the formula:
P = ρgh,
where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the height or depth.
In our case, the pressure at the interface is the pressure due to the sea water and the pressure due to the unknown liquid. The pressure at the interface due to the sea water can be calculated as:
P1 = ρ1gh1,
where ρ1 is the density of the sea water and h1 is the height of the sea water column.
Similarly, the pressure at the interface due to the unknown liquid can be calculated as:
P2 = ρ2gh2,
where ρ2 is the density of the unknown liquid and h2 is the height of the unknown liquid column.
Since the pressure at the interface is the same, we can equate the two expressions:
P1 = P2.
Substituting the expressions for P1 and P2, we get:
ρ1gh1 = ρ2gh2.
We are given ρ1 = 1.025 kg/L, h1 = 14.6 cm, and h2 = 1.0 cm.
Converting the densities to the same unit (in this case, g/cm³), we have:
ρ1 = 1.025 g/cm³ and ρ2 = ρ (unknown density of the liquid).
Substituting the values into the equation, we get:
1.025 g/cm³ * 14.6 cm = ρ * 1.0 cm.
Simplifying the equation, we have:
14.9295 g/cm² = ρ.
Therefore, the density of the unknown liquid is approximately 14.93 g/cm³.